document.write( "Question 1087707: You have a 25​-mile commute into work. Since you leave very​ early, the trip going to work is easier than the trip home. You can travel to work in the same time that it takes for you to make it 20 miles on the trip back home. Your average speed coming home is 2 miles per hour slower than your average speed going to work. What is your average speed going to​ work? \n" ); document.write( "
Algebra.Com's Answer #702027 by josgarithmetic(39620)\"\" \"About 
You can put this solution on YOUR website!
r, speed going
\n" ); document.write( "r-2, speed to go back home
\n" ); document.write( "t, the time going to work\r
\n" ); document.write( "\n" ); document.write( "\"system%28rt=25%2C%28r-2%29t=20%29\"\r
\n" ); document.write( "\n" ); document.write( "-
\n" ); document.write( "\"rt-2t=20\"
\n" ); document.write( "\"25-2t=20\", using substitution from \"going to\" equation
\n" ); document.write( "\"-2t=-5\"
\n" ); document.write( "\"t=5%2F2\"
\n" ); document.write( "-
\n" ); document.write( "\"rt=25\"
\n" ); document.write( "\"r=25%2Ft\"
\n" ); document.write( "\"r=25%2F%282.5%29\"
\n" ); document.write( "\"highlight%28r=10%29\"\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "---------\r
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\n" ); document.write( "\n" ); document.write( "
\r\n" );
document.write( "                RATE     TIME     DISTANCE\r\n" );
document.write( "GOING           r         t        rt=25\r\n" );
document.write( "BACK HOME       r-2       t         20\r\n" );
document.write( "
\r
\n" ); document.write( "\n" ); document.write( "Basic Travel Rate Rule, RT=D:
\n" ); document.write( "System, \"system%28rt=25%2C%28r-2%29%2At=20%29\"
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